$A$ cannon placed on a cliff at a height of $375 \ m$ fires a cannonball with a velocity of $100 \ ms^{-1}$ at an angle of $30^{\circ}$ above the horizontal. The horizontal distance between the cannon and the target is (Acceleration due to gravity $= 10 \ ms^{-2}$)

Two objects are located at a height of $10 \ m$ above the ground. At some point in time, the objects are thrown with an initial velocity of $2 \sqrt{2} \ m \ s^{-1}$ at an angle of $45^{\circ}$ and $135^{\circ}$ with the positive $X$-axis, respectively. Assuming $g = 10 \ m \ s^{-2}$, the velocity vectors will be perpendicular to each other at a time equal to: (in $s$)

Shots are fired from the top of a tower and from its bottom simultaneously at angles $30^o$ and $60^o$ as shown. If the horizontal distance of the point of collision is at a distance $a$ from the tower,then the height of the tower $h$ is:

$A$ child stands on the edge of a cliff $10 \ m$ above the ground and throws a stone horizontally with an initial speed of $5 \ m/s$. Neglecting air resistance,the speed with which the stone hits the ground will be $.......... \ m/s$ (given,$g = 10 \ m/s^2$).

$A$ body is projected from the top of a tower with a velocity $\overrightarrow{u} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \text{ m/s}$, where $\hat{i}, \hat{j},$ and $\hat{k}$ are unit vectors along east, north, and vertically upwards, respectively. If the height of the tower is $30 \text{ m}$, find the horizontal range of the body on the ground. (Take $g = 10 \text{ m/s}^2$) (in $\text{ m}$)

Two balls of mass $M$ and $2M$ are thrown horizontally with the same initial velocity $v_{0}$ from the top of a tall tower and experience a drag force of $-kv$ $(k > 0)$,where $v$ is the instantaneous velocity. Then,